# How do you evaluate e^( ( 13 pi)/8 i) - e^( ( pi)/4 i) using trigonometric functions?

Feb 16, 2016

${e}^{\frac{13 \pi}{8} i} - {e}^{\frac{\pi}{4} i} = \frac{\sqrt{2 - \sqrt{2}} - \sqrt{2}}{2} - i \cdot \frac{\sqrt{2 + \sqrt{2}} + \sqrt{2}}{2}$ ~=
$\cong - 0.3244 - 1.6310 \cdot i$

#### Explanation:

First of all, let's recall the Euler's formula that imaginary exponents:
${e}^{i \cdot x} = \cos \left(x\right) + i \cdot \sin \left(x\right)$

Let's evaluate two terms of the original expression separately and then determine the difference between them.

Based on this formula,
${e}^{\frac{13 \pi}{8} i} = \cos \left(\frac{13 \pi}{8}\right) + i \cdot \sin \left(\frac{13 \pi}{8}\right)$

Simple property of trigonometric functions that immediately follows from their definition are:
$\cos \left(x + \pi\right) = - \cos \left(x\right)$
$\sin \left(x + \pi\right) = - \sin \left(x\right)$
$\cos \left(x + \frac{\pi}{2}\right) = - \sin \left(x\right)$
$\sin \left(x + \frac{\pi}{2}\right) = \cos \left(x\right)$

Since $\frac{13 \pi}{8} = \pi + \frac{5 \pi}{8}$ and $\frac{5 \pi}{8} = \frac{\pi}{2} + \frac{\pi}{8}$, the expression above can be further evaluated to
$- \cos \left(\frac{5 \pi}{8}\right) - i \cdot \sin \left(\frac{5 \pi}{8}\right) = \sin \left(\frac{\pi}{8}\right) - i \cdot \cos \left(\frac{\pi}{8}\right)$

To determine $\sin \left(\frac{\pi}{8}\right)$ and $\cos \left(\frac{\pi}{8}\right)$, recall the formula for $\cos$ of double angle:
$\cos \left(2 x\right) = 2 {\cos}^{2} \left(x\right) - 1$
From it for $x = \frac{\pi}{8}$ follows:
${\cos}^{2} \left(\frac{\pi}{8}\right) = \frac{1}{2} \left(1 + \cos \left(\frac{\pi}{4}\right)\right) = \frac{1}{4} \left(2 + \sqrt{2}\right)$

$\cos \left(\frac{\pi}{8}\right) = \frac{1}{2} \sqrt{2 + \sqrt{2}}$

From ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ for $x = \frac{\pi}{8}$ follows
${\sin}^{2} \left(\frac{\pi}{8}\right) = 1 - \frac{1}{4} \left(2 + \sqrt{2}\right) = \frac{1}{4} \left(2 - \sqrt{2}\right)$

$\sin \left(\frac{\pi}{8}\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

So, the first term in the original expression is
${e}^{\frac{13 \pi}{8} i} = \frac{1}{2} \sqrt{2 - \sqrt{2}} - i \cdot \frac{1}{2} \sqrt{2 + \sqrt{2}}$

The second term in the original expression is
${e}^{\frac{\pi}{4} i} = \cos \left(\frac{\pi}{4}\right) + i \cdot \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + i \cdot \frac{\sqrt{2}}{2}$

The difference between the first and the second terms is:
$\frac{\sqrt{2 - \sqrt{2}} - \sqrt{2}}{2} - i \cdot \frac{\sqrt{2 + \sqrt{2}} + \sqrt{2}}{2}$